Which recursive formula can be used to define this sequence for n > 1? 11, 12, 13, 14, 15, 16, ... Choices: (A)a = 1a (B)a = 12/11a (C)a = a - 1 (D)a = a + 1

Sequence Type: We have the sequence: 11, 12, 13, 14, 15, 16, ... Is the given sequence geometric or arithmetic? The difference between consecutive terms is the same. The given sequence is arithmetic.Find Common Difference: Find the common difference, d. Two consecutive terms are 11 and 12. 12 - 11 = 1 Common difference (d): 1Recursive Formula: Identify the recursive formula for the given sequence. Since the common difference is 1, the recursive formula will be of the form a_n = a_(n-1) + d. Substitute 1 for d in a_n = a_(n-1) + d. Recursive formula: a_n = a_(n-1) + 1

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Which recursive formula can be used to define this sequence for n > 1? $\newline$ $11, 12, 13, 14, 15, 16, \ldots$ $\newline$ Choices: $\newline$ (A) $a_n = 1a_{n-1}$ $\newline$ (B) $a_n = \frac{12}{11}a_{n-1}$ $\newline$ (C) $a_n = a_{n-1} - 1$ $\newline$ (D) $a_n = a_{n-1} + 1$

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Math Problems

Algebra 1

Write a formula for a recursive sequence

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Q. Which recursive formula can be used to define this sequence for

n > 1

\newline

11, 12, 13, 14, 15, 16, \ldots

\newline

Choices:

\newline

(A)

a_n = 1a_{n-1}

\newline

(B)

a_n = \frac{12}{11}a_{n-1}

\newline

(C)

a_n = a_{n-1} - 1

\newline

(D)

a_n = a_{n-1} + 1

Sequence Type: We have the sequence: $11, 12, 13, 14, 15, 16, \ldots$ $\newline$ Is the given sequence geometric or arithmetic? $\newline$ The difference between consecutive terms is the same. $\newline$ The given sequence is arithmetic.
Find Common Difference: Find the common difference, $d$ . $\newline$ Two consecutive terms are $11$ and $12$ . $\newline$ $12 - 11 = 1$ $\newline$ Common difference ( $d$ ): $1$
Recursive Formula: Identify the recursive formula for the given sequence. $\newline$ Since the common difference is $1$ , the recursive formula will be of the form $a_n = a_{(n-1)} + d$ . $\newline$ Substitute $1$ for $d$ in $a_n = a_{(n-1)} + d$ . $\newline$ Recursive formula: $a_n = a_{(n-1)} + 1$